# Degree $3$ complex polynomial with real coefficients have $3$ complex roots (without FTA)

Let $$f(z)$$ be a degree $$3$$ complex polynomial with real coefficients. How do we prove that $$f(z)=0$$ has three complex solutions without using the Fundamental Theorem of Algebra.

If $$f(x)$$ was real variable polynomial, I could use the Intermediate value theorem to at least get one root but with complex variable I have no idea what to do or how to start.

• You can solve the cubic by radicals using your favourite method - which is algebraically identical to the solution of a cubic with real coefficients – Mark Bennet Oct 28 '18 at 18:20

Let $$f(z)$$ be a degree $$3$$ complex polynomial with real coefficients.
... tells you precisely that you can think of $$f$$ as a function $$f\colon\mathbb R\to\mathbb R$$ and can use intermediate value theorem.
After you obtain one real root $$\alpha$$, divide $$f$$ by $$x-\alpha$$ to get a degree $$2$$ polynomial. I assume you know how to prove how many roots it has.
Then by dividing your polynomial by $$x- \alpha$$ you find the other two solutions, which may be real or complex.